Question

If ax2/3 + bx1/3 + c = 0 then the value of a3x2 + b3x + c3 is given by​

Answer 1

Topic

Polynomials

Solution

We observe that,

$a^{3}x^{2}+b^{3}x+c^{3}=(ax^{\frac{2}{3} })^{3}+(bx^{\frac{1}{3} })^{3}+(c)^{3}$

and,

$ax^{\frac{2}{3} }+bx^{\frac{1}{3} }+c=0$.

We are going to use the fact that,

• $a^{3}+b^{3}+c^{3}=3abc$ if $a+b+c=0$ or $a=b=c\in \mathbb{R}$. $\bold{[1]}$

The calculation is as follows.

$(ax^{\frac{2}{3} })^{3}+(bx^{\frac{1}{3} })^{3}+(c)^{3}=3\cdot ax^{\frac{2}{3} }\cdot bx^{\frac{1}{3} }\cdot c$

$\implies (ax^{\frac{2}{3} })^{3}+(bx^{\frac{1}{3} })^{3}+(c)^{3}=\boxed{3abcx}$

More information

$\bold{[1]}$ The identity $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=a^{3}+b^{3}+c^{3}-3abc$ states the value of $a^{3}+b^{3}+c^{3}-3abc$ is zero when $a+b+c=0$. Furthermore, we see that $a^{2}+b^{2}+c^{2}-ab-bc-ca=\dfrac{1}{2} \{(a-b)^{2}+(b-c)^{2}+(c-a)^{3}\}$, if $a,b,c$ are real numbers, the squares of real numbers are restricted to zero or positive. So, the second factor becomes zero when $a=b=c$.